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Ksenya-84 [330]
2 years ago
12

A 1.50-liter sample of dry air in a cylinder exerts a pressure of 3.00 atmospheres at a temperature of 25°C. Without changing th

e temperature, a piston is moved in the cylinder until the pressure in the cylinder is reduced to 1.00 atmospheres. The volume of the gas is _____.

Chemistry
2 answers:
Aneli [31]2 years ago
3 0
Hope this helps you.

Galina-37 [17]2 years ago
3 0

Answer:

V_2=4.5 L

Explanation:

Hi, air can be considered as an ideal gas. If mantaining constant the temeperature, you can apply the Boyle-Mariotte formula:

P_1*V_1=P_2*V_2

In this case:

P_1=3 atm

V_1=1.5 L

P_2=1 atm

So:

V_2=\frac{P_1*V_1}{P_2}

V_2=\frac{3atm*1.5 L}{1atm}

V_2=4.5 L

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A voltaic cell consists of a nickel electrode in a solution containing Ni2+ ions, and a copper electrode in a solution containin
Marysya12 [62]
First, we write the half equations for the reduction of the chemical species present:

Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V

In order to determine the potential of the cell, we find the difference between the two values. For this:

E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V

The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
4 0
1 year ago
You carefully weigh out 13.00 g of CaCO3 powder and add it to 52.65 g of HCl solution. You notice bubbles as a reaction takes pl
Sedbober [7]
CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O

n(CaCO₃)=m(CaCO₃)/M(CaCO₃)
n(CaCO₃)=13.00/100.09=0.1299 mol

Δm=13.00+52.65-60.32=5.33 g

m(CO₂)=5.33 g
n(CO₂)=5.33/44.01=0,1211 mol

w=0.1211/0.1299=0,9323 (93.23%)

7 0
2 years ago
Read 2 more answers
If 25 g of NH3, and 96 g of H2S react according to the following reaction, what is the
jeyben [28]

25 g of NH₃ will produce 47.8 g of (NH₄)₂S​

<u>Explanation:</u>

2 NH₃ + H₂S ----> (NH₄)₂S​

Molecular weight of NH₃ = 17 g/mol

Molecular weight of (NH₄)₂S​ = 68 g/mol

According to the balanced reaction:

2 X 17 g of NH₃ produces 68 g of (NH₄)₂S​

1 g of NH₃ will produce \frac{68}{34} g of (NH₄)₂S​

25g of NH₃ will produce \frac{65}{34} X 25 g of (NH₄)₂S​

                                     = 47.8 g of (NH₄)₂S​

Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S​

4 0
2 years ago
You have 125 g of a certain seasoning and are told that it contains 62.0 g of salt. what is the percentage of salt by mass in th
r-ruslan [8.4K]
62.0g/125g= 0.496 x 100 = 49.6%
4 0
2 years ago
What is the mass of a sample of NH3 containing 7.20 × 1024 molecules of NH3? 161 grams 187 grams 203 grams 214 grams
Shkiper50 [21]

Answer:

203 grams

Explanation:

<em>It is known that 1.0 mole of a compound contains Avogadro's number of molecules (6.022 x 10²³). </em>

<em><u>Using cross multiplication:</u></em>

1.0 mol contains → 6.022 x 10²³ molecules.

??? mol contains → 7.2 x 10²⁴ molecules.

∴ The no. of moles of (6.3 x 10²⁴ molecules) of NH₃ = (1.0 mol)(7.2 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 11.96 mol.

<em>∴ The no. of grams of NH₃ present = no. of moles x molar mass </em>= (11.96 mol)(17.0 g/mol) = <em>203.3 g ≅ 203.0 g.</em>

7 0
2 years ago
Read 2 more answers
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